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Fixed subgroups of automorphisms of hyperbolic 3-manifold groups
Topology and its Applications 173 (2014) 175–187
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Topology and its Applications www.elsevier.com/locate/topol
Fixed subgroups of automorphisms of hyperbolic 3-manifold groups Jianfeng Lin, Shicheng Wang ∗ Department of Mathematics, Peking University, Beijing, China
a r t i c l e
i n f o
Article history: Received 18 February 2014 Accepted 6 May 2014 Available online xxxx
a b s t r a c t For fixed subgroups Fix(φ) of automorphisms φ of hyperbolic 3-manifold groups π1 (M ), we observe that rk(Fix(φ)) < 2 rk(π1 (M )) and the constant 2 in the inequality is sharp; we also classify all possible groups Fix(φ). © 2014 Elsevier B.V. All rights reserved.
Keywords: Fixed subgroups Automorphisms Hyperbolic 3-manifold
1. Introduction For a group G and an automorphism φ : G → G, we define Fix(φ) = {ω ∈ G | φ(ω) = ω}, which is a subgroup of G, and use rk(G) to denote the rank of G. The so called Scott conjecture proved 20 years ago in a celebrate work of M. Bestvina and M. Handel [1] states that: Theorem 1.1. For each automorphism φ of a free group G = Fn , rk Fix(φ) ≤ rk(G). In a recent paper by B.J. Jiang, S.D. Wang and Q. Zhang [4], it is proved that Theorem 1.2. For each automorphisms φ of a compact surface group G = π1 (S), rk Fix(φ) ≤ rk(G). It is obvious that the bounds given in Theorem 1.1 and Theorem 1.2 are sharp and can be achieved by the identity maps. * Corresponding author. E-mail addresses: [email protected] (J. Lin), [email protected] (S. Wang). http://dx.doi.org/10.1016/j.topol.2014.05.020 0166-8641/© 2014 Elsevier B.V. All rights reserved.
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In this note, we will address the similar problem for hyperbolic 3-manifold groups. We call a compact 3-manifold M hyperbolic, if M is orientable, and the interior of M admits a complete hyperbolic structure of finite volume (then M is either closed or ∂M is a union of tori). In this case G = π1 (M ) is isomorphic to a cofinite volume torsion free Kleinian group. A main observation in this paper is the following: Theorem 1.3. For each automorphism φ of a hyperbolic 3-manifold group, G = π1 (M ), rk Fix(φ) < 2 rk(G), and the upper bound is sharp when G runs over all hyperbolic 3-manifold groups. Theorem 1.3 is a conclusion of the following Theorem 1.4 and Theorem 1.5. Theorem 1.4. There exists a sequence of automorphisms φn : π1 (Mn ) → π1 (Mn ) of closed hyperbolic 3-manifolds Mn such that Fix(φn ) is the group of a closed surface, and rk(Fix(φn )) >2− rk(π1 (Mn ))
as n → ∞
for any > 0. Theorem 1.5. Suppose φ is an automorphism of G = π1 (M ), where M is a hyperbolic 3-manifold. Then rk(Fix(φ)) < 2 rk(G). The proof of Theorem 1.4 is self-contained up to some primary (and elegant) facts on hyperbolic geometry and on combinatoric topology and group theory. Roughly speaking each (Mi , φi ) in Theorem 1.4 is constructed as follows: We first construct the hyperbolic 3-manifold Pi with connected totally geodesic boundary. Then we double two copies of Pi along their boundaries to get the closed hyperbolic 3-manifold Mi . The reflection of Mi alone ∂Pi induces an automorphism φi : π1 (Mi ) → π1 (Mi ) with Fix(φ) = π1 (∂Pi ). In this process all involved ranks are carefully controlled, we get the inequality in Theorem 1.4. To prove Theorem 1.5, besides some combinatorial arguments on topology and on group theory, we need the following Theorem 1.6 which classifies all possible groups Fix(φ) for automorphisms φ of hyperbolic 3-manifold groups. Recall that each automorphism φ of π1 (M ) can be realized by an isometry f on M according to Mostow rigidity theorem. Theorem 1.6. Suppose G = π1 (M ), where M is a hyperbolic 3-manifold, and φ is an automorphism of G. Then Fix(φ) is one of the following types: the whole group G; the trivial group {e}; Z; Z ⊕ Z; a surfaces group π1 (S), where S can be orientable or not, and closed or not. More precisely (1) Suppose φ is induced by an orientation preserving isometry. (i) Fix(φ) is either Z, or Z ⊕ Z, or G, or {e}; moreover (ii) If M is closed, then Fix(φ) is either Z or G; (2) Suppose φ is induced by an orientation reversing isometry f . (i) If φ2 = id, then Fix(φ) is either Z or {e}; (ii) If φ2 = id, then Fix(φ) is either {e}, or the surface group π1 (S), where S is an embedded surface in M that is pointwise fixed by f . Theorem 1.6 is proved by using the algebraic version of Mostow Rigidity theorem, as well as some hyperbolic geometry and covering space argument.
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Fig. 1. Truncated hyperbolic tetrahedron.
The paper is organized as follows: In Section 2 we will prove Theorem 1.4 generalizing the examples from closed hyperbolic 3-manifolds to hyperbolic 3-manifolds with cusps. Theorem 1.6 and Theorem 1.5 will be proved in Section 3 and Section 4 respectively. Suppose a compact 3-manifold M is hyperbolic and S is a proper embedded surface in M . We say S is totally geodesic surface if S o , the interior of S, is totally geodesic. With an abuse of notation, we will use the same notation M (S) to present the interior of M (S). For terminologies not defined, see [3] and [10] for geometry and topology of 3-manifolds, and see [9] for group theory. 2. Proof of Theorem 1.4 In this section, we construct examples stated in Theorem 1.4. As sketched in Section 1, the first step is to construct hyperbolic manifolds with boundaries and control the ranks of the fundamental groups. There are different approaches to construct such manifolds. We will use the most original and the most direct one due to Thurston. (For another approach see Remark 2.6.) In Thurston’s Lecture Notes (Section 3.2 of [10]), there is a very concrete and beautiful construction of hyperbolic 3-manifolds with totally geodesic boundaries involving primary hyperbolic geometry only. In 3-dimensional hyperbolic space H 3 , there is a one-parameter family of truncated hyperbolic tetrahedron as in Fig. 1: Each of its 8 faces is totally geodesic; each of its 18 edges is geodesic line segment. There are 4 triangle faces and 4 hexagon faces. The 12 edges of the 4 triangle faces have the same length, and the remaining 6 edges, we call them “inner edge”, also have the same length. The triangle faces are perpendicular to the hexagon faces. The angles between hexagon faces are all equal and can be arbitrary angles between 0◦ and 60◦ . We will use those polyhedra to construct hyperbolic manifolds with totally geodesic boundaries. Suppose we have some copies of tetrahedron. We pair the faces of tetrahedron and glue them together (therefore some edges and vertexes are also glued together). After that, we remove a neighborhood of the vertex and get a topological manifold P . A tetrahedron with its vertex neighborhood removed is homeomorphic to the truncated simplex mentioned above. Now suppose for some fixed k > 6, all k edges of the tetrahedron are glued together in the resulting complex. We can set the face angle α of the truncated simplex to be 2π k such that the hyperbolic structure of the truncated simplex fix together to give the hyperbolic structure of P , and the triangle faces of the truncated simplex are matched together to form the totally geodesic ∂P . It is easy to see that the number of vertexes of tetrahedrons (after gluing) equals the number of the boundary components.
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Fig. 2. New generators.
Moreover, if we remove the regular neighborhood of the inner edges in P , we will get a handlebody H. To see this, we remove the neighborhood of the 6 edges of a tetrahedron. Topologically, it is homeomorphic to D3 and the 4 tetrahedron faces are 4 disjoint disks on ∂D3 . Then, we glue them together. If we glue some 3-balls along disks on their boundaries, we get a handlebody. Therefore, P can be obtained by attaching m 2-handles on a handlebody of genus n + 1. It is easy to see that m is the number of inner edges after gluing and n is the number of tetrahedrons. Now we double P along its boundary to get a closed hyperbolic manifold DP. We have to control the rank of π1 (DP). This is done in the following lemma. Lemma 2.1. Suppose P is obtained by attaching l 2-handles to a handlebody of genus k. Then rk(π1 (DP)) ≤ k + l, where DP is the double of P . Proof. Suppose P is obtained from a handlebody H of genus k by attaching l 2-handles h1 , h2 , ..., hl with attaching curves γ1 , γ2 , ..., γl , where γ1 , γ2 , ..., γl are disjoint simple closed curves on ∂H, and for each 2-handle D2 × I, ∂D × I is identified with the attaching region N (γi ), the regular neighborhood of γi , for some i. Then we have P =H
{hi }.
{N (γi )}
Note that in the doubling DP, the two copies of handlebody H and H are glued together along ∂H − D2 × I of the 2-handle hi are glued alone the D2 × ∂I to get a solid torus i N (γi ), and each two copies Si , which is attached to H ∂H− N (γi ) H along the torus boundary formed by two copies of N (γi ) (see i Fig. 2). We have
DP =
H
∂H− i N (γi )
H
Si .
i
Because attaching solid tori along the torus boundaries of H ∂H− N (γi ) H does not increase the rank of i the fundamental group, we just need to control rk(π1 (H ∂H− N (γi ) H )). We consider the two skeleton of i this space. The two skeleton of the handlebody H consists of a surface of genus k and k copies of compressing disks. Therefore, the two-skeletons of H ∂H− N (γi ) H can be obtained as follows: starting from a surface i Sk of genus k, we glue two copies of Sk along Sk − i N (γi ). This is equal to attaching l copies of annulus along ∂N (γi ), i = 1, 2, ..., l. Then we glue k compressing disks on both sides.
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Fig. 3. Gluing tetrahedrons.
Comparing to the two-skeleton of H, we see that only l new generators are involved by the attached annulus and the new attaching disks do not increase the rank of fundamental group. Thus we get rk(π1 (H ∂H− N (γi ) H )) ≤ k + l. 2 i
Now we can construct our examples. We start from n (n > 3, 3 n) copies of the tetrahedron indicated in Fig. 3, where the edges are marked. We represent the faces by the edges around it. Each tetrahedron Ti has 4 faces (1, 3, 2)i , (4, 5, 3)i , (2, 6, 4)i , (5, 1, 6)i . Then we group the 4n faces into 2n pairs:
(1, 3, 2)i , (4, 5, 3)i+1 ;
(2, 6, 4)i , (5, 1, 6)i+1 ,
i = 1, 2, ..., n, and n + 1 ≡ 1.
The two faces in each pair are glued together, and the orders of the edges are preserved. (It’s easy to see that the arrows on the edges are preserved too.) Then we get a simplicial complex X. We write ai ↔ bj to indicate that the edge a in Ti is glued together with b in Tj . With this notation, we have 1k ↔ 4k+1 ↔ 6k+2 ↔ 1k+3 ;
2k ↔ 5k+1 ↔ 3k ↔ 2k−1 .
We first count the number of the edges after the gluing: Since 3 n, we see that the 3n edges 1∗ , 4∗ , 6∗ are glued together, and the 3n edges 2∗ , 5∗ , 3∗ are glued together, so there are two edges in X. Then we count the number of the vertices after the gluing: If we denote the initial point and the terminal point of the directed edge ik by I(ik ) and E(ik ), then we have: E(1k+1 ) ↔ I(3k+1 ) ↔ I(2k ) ↔ I(4k ) ↔ I(1k−1 ) ↔ E(2k−1 )
(2.1)
The first, third and fifth identifications are shown in Fig. 3. The second and fourth identifications follow from the facts that 3k+1 and 2k , 4k and 1k−1 are glued together as direct edges respectively. Since the two edges [1∗ ], [2∗ ] after the gluing. Eq. (2.1) implies that all ends of [1∗ ], [2∗ ] are identified to a point. Hence there is only one vertex in the simplicial complex X. Finally we check the orientation: If we use the right hand coordinate system, the faces (1, 3, 2) and (5, 1, 6) correspond to outward normal vectors while (2, 6, 4) and (4, 5, 3) correspond to inward ones. Since each inward face is glued with an outward one, the orientations are matched after gluing. Now if we remove a regular neighborhood of the unique vertex, by the discussion at the begin of this section, we get an orientable hyperbolic three manifold P with connected, totally geodesic boundary. As we have discussed, P can be constructed by attaching two 2-handles on a handle body of genus n + 1. The genus of ∂P is n − 1. Now double P along its boundary to get a closed 3-manifold DP. If we
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choose a point p ∈ ∂P as the base point, the reflection f on DP along ∂P will induce an automorphism φ : π1 (DP) → π1 (DP). The following lemma is a general fact about automorphisms that switch the two factors of the amalgamation of a group with itself. Lemma 2.2. In the construction above, Fix(φ) = Im(i∗ (π1 (∂P ))). Proof. Let G and G be two identical copies of π1 (P ) and H be their common subgroup π1 (∂P ) (recall that P is boundary incompressible). Then we have: π1 (DP) = G ∗H ∗G . For each right coset gi H of H in G, fix its representative gi . We choose the unit 1 as the representative for the right coset H itself. Then {gi H} give the right coset decomposition of H in G . According to [9, Theorem 1.7], each element γ in G ∗H ∗G can be written uniquely in a form γ = a1 b1 a2 b2 ...an bn h, where h ∈ H, ai is some representative gj and bi is some representative gk ; moreover ai = 1 implies i = 1 and bi = 1 implies i = n. Then φ(γ) = a1 b1 a2 b2 ...an bn h. By uniqueness of the standard form, it is direct to see that if φ(γ) = γ, then γ = h. hence only elements in H can be fixed by φ. 2 By Lemma 2.1, rk(π1 DP) ≤ n + 3. Since ∂P has genus n − 1, by Lemma 2.2 Fix(φ) = Im(i∗ (π1 (∂P ))) ∼ = π1 (∂P ) has rank 2n − 2. For each n > 2, construct such pair (DP, φ), and denoted as (Mn , φn ). Then 2n − 2 rk(Fix(φn )) ≥ > 2 − , rk(π1 (Mn )) n+3 for any > 0. Hence we finished the proof of Theorem 1.4.
as n → ∞
2
The construction in Theorem 1.4 for closed hyperbolic 3-manifold can be modified to the case of hyperbolic 3-manifold with cusps. Precisely: Proposition 2.3. There exists a sequence of hyperbolic 3-manifolds Mn with cusps and automorphisms φn : π1 (Mn ) → π1 (Mn ), such that Fix(φn ) is a free group, and rk(Fix(φn )) >2− rk(π1 (Mn ))
as n → ∞.
for any > 0. We give some theorems to prove Proposition 2.3. Theorem 2.4. ([11]) Suppose M is a hyperbolic 3 manifold with finite volume and f is an involution of M . Than M admits a hyperbolic structure with finite volume such that f is an isometry with respect to this structure. Theorem 2.5. ([5,12]) Suppose M is a hyperbolic 3-manifold with finite volume and α is a simple closed geodesic in M . Then M − α admits a hyperbolic structure with finite volume.
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Proof. In the proof of Theorem 1.4, the hyperbolic 3-manifold P with connected totally geodesic boundary is obtained by attaching two 2-handles to a handlebody H along the attaching curve γ1 and γ2 . Now we choose a simple non-separating closed geodesic α in ∂P such that α ⊂ ∂H \ N (γ1 ) ∪ N (γ2 ). Then α remains a geodesic in DP. Remove α from the closed hyperbolic manifold DP, we get a new hyperbolic manifold with a cusp by Theorem 2.5, denoted by DP . The reflection f on DP along ∂P defines an involution f on DP . By Theorem 2.4, DP admits a hyperbolic structure so that f is an isometry under this hyperbolic structure. As the fixed point set of an isometry, the non-closed surface ∂P − α must be totally geodesic, and therefore incompressible. If we chose the base point to be on ∂P − α and consider the automorphism φ induced by f , the same combinational group theory argument as before shows that Fix(φ) = π1 (∂P − α). Because ∂P has genus n − 1, n is the same as in the proof of Theorem 1.4, ∂P − α is two punctured surface of genus n − 2, hence π1 (∂P − α) is a free group of rank 2(n − 2) + 1 = 2n − 3. Now we control rk(π1 (DP )) via the same technique in the proof of Lemma 2.1: DP consist of two parts: the first part consists of two copies of the handlebody H glued along ∂H \ N (γ1 ) ∪ N (γ2 ) ∪ N (α); the second part consists of two solid tori resulting from the doubling of the 2-handles. The same argument as in the proof of Lemma 2.1 shows that rk(π1 (DP )) < n + 4, and we have 2n − 3 rk(Fix(φ)) > 2 − , ≥ n+4 rk(π1 (DP ))
as n → ∞,
for any > 0. 2 Remark 2.6. There is another way to construct Mn which we sketch as follows. It is based on a result on Heegaard Splitting: Theorem 2.7. ([2]) Let M be a closed oriented 3-manifold which is Seifert fibered or which contains an essential torus. Then any splitting of M is a Heegaard distance ≤ 2 splitting. Note that Theorem 2.7 in [2] is stated for closed 3-manifolds, but the argument there can be used to prove the similar theorem for non-closed case. Now consider a manifold M obtained by attaching a two handle to a handlebody of genus n (n > 1). Suppose that the distances in the curve complex between the attaching curve and the boundaries of the compressing disks of the handlebody are larger than 3. By [2], M contains no essential surface of genus smaller than 2. By Haken hyperbolization theorem in [11], M is a hyperbolic manifold with totally geodesic boundary. Let Mn be the double of M . By Lemma 2.1, the inequality in Theorem 1.4 can be proved as before. 3. Proof of Theorem 1.6 In this section, we will classify all the possible fixed subgroups of automorphisms of torsion free cofinite volume Kleinian groups. We use Iso(H3 ) (resp. Iso+ (H3 )) to denote the group of (resp. orientation preserving) isometries of the 3-dimensional hyperbolic 3-space. The most important tool is the following algebraic version of Mostow rigidity theorem. Most topologists know the geometric version of Mostow rigidity: Any homotopy equivalence between finite volume hyperbolic 3 manifolds can be homotopied to an isometry. The following algebraic version appears in [7], which is equivalent to the geometric version. Theorem 3.1. Let Γ1 and Γ2 be two cofinite volume Kleinian groups, and φ : Γ1 → Γ2 be an isomorphism between them. Then there exists γ ∈ Iso(H3 ) (γ may be orientation reversing) such that φ(α) = γαγ −1 for any α ∈ Γ1 .
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Now let’s prove Theorem 1.6. Proof. Since G = π1 (M ) is a hyperbolic 3-manifold group, G can be considered as cofinite volume torsion free Kleinian group in Iso+ (H3 ). Now G acts on H3 as the deck transformation group for the covering π : H3 → H3 /G ∼ = M . By Theorem 3.1, there exists γ ∈ Iso(H3 ) such that for any α ∈ G, φ(α) = γαγ −1 . Then Fix(φ) = {α ∈ G | αγ = γα}.
(3.1)
Because γGγ −1 = G, γ induces an isometry f of M such that the following diagram commutes. H3
γ
π
M
H3 π
f
(3.2)
M
The verification of Theorem 1.6 will be based on (3.1) and the verification will be divided into two cases according to whether γ in (3.1) is orientation preserving or not. Case (1): γ is orientation preserving. (i) If γ = e then clearly Fix(φ) = G. Below we assume that γ is nontrivial. It is well-known that each element in G is either hyperbolic or parabolic; moreover two nontrivial elements α, β in Iso+ (H3 ) commute if and only if one of the following cases happens: (a) Both α and β are parabolic elements and they share the same fixed point in the infinite sphere S ∞ . (b) Both α and β are non-parabolic elements (elliptic or hyperbolic) and they share the same axis. (c) Both α and β are elliptic elements with rotation angle π and their axes are perpendicular to each other. Since elements in G (therefore in Fix(φ)) cannot be elliptic, we just need to consider case (a) and case (b). In these two cases, if α, β, γ are all nontrivial, α commutes with β, β commutes with γ, then α commute with γ. We see that Fix(φ) is a torsion free abelian group. As we know, the fundamental group of a hyperbolic 3-manifold can contain torsion free abelian subgroups of ranks at most 2. We have proved that: if φ is induced by an orientation preserving map, Fix(φ) can only be e, Z, Z ⊕ Z, or G. (ii) If we further assume that M is closed, then we have more restrictions. First π1 (M ) contains no subgroup Z ⊕ Z for a closed hyperbolic 3-manifold M . Also we claim that Fix(φ) = {e}. In fact, the self-isometry of a closed hyperbolic 3 manifold is always periodic. So there exists positive integer n such that f n = id. By commuting diagram (3.2), γ n induces the identity on M , therefore γ n ∈ G. Then clearly γ n ∈ Fix(φ). If γ n = e, then Fix(φ) is not trivial. If γ n = e, then γ is an elliptic element, so there is an axis l pointwise fixed by γ. By the commuting diagram (3.2), π(l) is pointwise fixed by the isometry f . Since M is closed, the fixed point set of an orientation preserving isometry can only be a closed geodesic. So π(l) is a closed geodesic. This means that there is a hyperbolic covering transformation α ∈ G sharing the axis l with γ, so αγ = γα, and therefore α ∈ Fix(φ) by (3.1). We have actually proved that if φ is induced by an orientation preserving map and M is closed, then Fix(φ) can be either Z or G. Case (2): γ is orientation reversing. Note Fix(φ) ⊆ Fix(φ2 ) and φ2 is induced by an orientation preserving map. There are two subcases now: (i) φ2 = id. Then Fix(φ2 ) can only be e, Z or Z ⊕ Z by Case (1)(i) and its proof, therefore Fix(φ) can only be e, Z or Z ⊕ Z.
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But in fact the situation Z ⊕ Z never happens. Because in this situation Fix(φ2 ) = Z ⊕ Z, which was generated by two parabolic elements β1 and β2 sharing the same fixed point p on the infinite sphere. Since we assume that φ2 = id, γ 2 is also a parabolic element with the fixed point p. Since Fix(γ) ⊂ Fix(γ 2 ), one can derive that γ has the unique fixed point p in H3 ∪ S ∞ , and in the upper-half model of H3 (we set p = ∞), γ 2 , β1 , β2 are translations along some directions v1 , v2 , v respectively. Consider their extended action on the plane z = 0. Then γ acts as a conformal (orientation reversing) map on this plane (in order to see this, we can just compose γ with an arbitrary reflection r which fixes p to get an orientation preserving isometry. By classical fact, both r ◦ γ and r act conformally on plane z = 0, then so does γ). Because γ 2 is a translation, γ must act as an orientation reversing isometry on the plane z = 0. So γ|z=0 = r ◦ h, where h is a translation along the direction v and r a reflection along an invariant line of h. Then it is a direct verification that γ commutes with βi if and only if the directions of v and vi are either the same or opposite. But the directions of v1 and v2 are neither the same nor opposite, so γ cannot commute with both two generators of Z ⊕ Z. We have proved that in this subcase Fix(φ) is either e or Z. (ii) φ2 = id. Then γ 2 commute with the whole group G. Thus γ 2 = e and γ has order 2. An order 2 orientation reversing isometry of H 3 can only be the reflection along a single point or reflection along a geodesic plane. If γ is the reflection along a point p ∈ H3 , then p is the only fixed point of γ. For any α ∈ Fix(φ), we have γα = αγ by (3.1). Hence γα(p) = αγ(p) = α(p), that is γ also has fixed point α(p), hence p = α(p). Because α is a covering transformation, we must have α = e. We have proved that if γ is a reflection along a single point, then Fix(φ) is trivial. If γ is the reflection along a totally geodesic plane P . Then P is pointwise fixed by γ. Because of the commuting diagram (3.2), π(P ) is pointwise fixed by f . We know that the fixed point set of an orientation reversing isometry of a hyperbolic 3-manifold must be totally geodesic surfaces if it is dimension 2. Thus π(P ) S where S is a totally geodesic surface in M . (S may be non-orientable although M is orientable. And if M has cusps, S may have cusps too.) For any α ∈ Fix(φ), γα = αγ. Then for each x ∈ P , γα(x) = αγ(x) = α(x), that is α(x) ∈ P . It follows that P is invariant under α. Conversely, suppose a covering transformation α ∈ G such that α(P ) = P . Then it is easy to see that γα = αγ and therefore α ∈ Fix(φ). So Fix(φ) is exactly the covering transformations of the universal π|P covering map P −−−→ S. We have proved that Fix(φ) ∼ = π1 (S). 2 4. Proof of Theorem 1.5 By Theorem 1.6, the situation rk(Fix(φ)) > rk(π1 (M )) can appear only if Case (2)(ii) in Theorem 1.6 happens, and if Case (2)(ii) in Theorem 1.6 happens, then Fix(φ) = π1 (S) for some surface S which is pointwise fixed by an orientation reversing isometry f of order 2 on M . The proof of Theorem 1.5 will be completed by the following: Proposition 4.1. Suppose M is a hyperbolic 3-manifold and S is a proper embedded surface in M . If there is an orientation reversing isometry f of order 2 on M fixing S pointwisely. Then rk π1 (S) < 2 rk π1 (M ) . To prove Proposition 4.1, we need the following lemma which contains several elementary facts: Lemma 4.2. (1) Suppose G is a group with subgroup H of index n. Then rk(G) ≥
rk(H) + n − 1 . n
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(2) Suppose S is a boundary component of the compact 3-manifold M , and DS (M ) is the doubling of two copies of M along S. Then rk π1 DS (M ) ≥ rk π1 (M ) . (3) (Half lives, half dies Lemma) Suppose M is a compact orientable 3-manifold. Then
dim H 1 (∂M, Q) dim image i∗ : H 1 (∂M, Q) → H 1 (M, Q) = 2 where i∗ is induced by the inclusion i : ∂M → M . (4) Suppose M is a compact orientable 3-manifold and S is an incompressible boundary component of M . If the homomorphism π1 (S) → π1 (M ) induced by the inclusion is not a surjection, then there is a finite ˜ → M such that p−1 (S) contains more then one component. covering p : M Proof. (1) If rk(G) = k, we can find a 2-dimensional CW complex X with fundamental group G so that X ˜ be the n-sheet covering space corresponding to the subgroup H. has only one vertex and k edges. Let X ˜ and this lifted CW complex of X ˜ provides a presentation of Then there are n vertexes and nk edges in X, H with n(k − 1) + 1 generators. So rk(H) ≤ n(k − 1) + 1, that is rk(G) = k ≥
rk(H) + n − 1 . n
(2) It is clear that there is a reflection f about S on DS (M ), which provides a folding map DS (M ) → DS (M )/f ∼ = M , and which obviously induce an epimorphism between fundamental groups. Hence rk π1 (M ) ≥ rk π1 (M1 ) . (3) See [8, Section 23]. (4) By the assumptions, π1 (S) is a proper subgroup of π1 (M ) (we pick a base point x on S for both π1 (S) and π1 (M )). Pick a non-zero element α of π1 (M ) but not in π1 (S). By [6, Theorem 1] (peripheral subgroups are separable), there is a finite index subgroup H of π1 (M ) which contains π1 (S) but does not ˜ → M corresponding to H, then there is a component S˜ of contain α. Consider the finite covering p : M −1 p (S) homeomorphic to S (since π1 (S) ⊂ H), and p−1 (S) has more than one component (since α does not in H, the lift α ˜ of α with one end in S˜ must have another end in another component of p−1 (S)). 2 Lemma 4.3. Suppose M is a compact orientable 3-manifold and S is an incompressible boundary component of M with genus g. Then rk(π1 (M )) > g. Proof. Since S is incompressible, the homomorphism i∗ : π1 (S) → π1 (M ) induced by the inclusion is injective. Suppose i∗ is also surjective. Then rk(π1 (M )) = 2g and the inequality is trivial. Now assume i∗ is not surjective. Let p : M → M be the n sheet covering provided by the proof of Lemma 4.2(4). Then the preimage of S has m > 1 component. Now we can compute the sum of the genus of these m boundary components. Because S has Euler number 2 − 2g, the sum of the Euler number of the preimage of S is n(2 − 2g). Therefore, the sum of their genus is n(g − 1) + m. Since H 1 (∂M , Q) is a direct sum of the homology of the boundary components, it contains at least 2n(g − 1) + 2m copies of Q. By Lemma 4.2(3), we have rk(H 1 (M , Q)) ≥ n(g −1) +m. Since H 1 (M , Q) is a quotient group of π1 (M ), we have rk(π1 (M )) ≥ n(g − 1) + m. Since π1 (M ) is subgroup of π1 (M ) of index n, by Lemma 4.1(1) we have n(g − 1) + m + n − 1 m−1 =g+ . rk π1 (M ) ≥ n n Since m > 1. Then Lemma 4.3 is proved. 2
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Remark 4.4. Suppose M is a closed hyperbolic 3-manifold and S is a closed surface embedded in M which is pointwise fixed by an orientation reversing involution on M . Cutting M along S, we get a compact 3-manifold M (may be not connected) with a boundary component S. In the proof of Theorem 4.1 we will apply Lemma 4.3 to S ⊂ M directly by the following reason: Using the fact that S is a proper embedded surface in M which is pointwise fixed by an orientation reversing involution f on M and M contains no essential spheres and essential tori, it follows that S must be boundary incompressible (otherwise M would contain essential spheres). Now we start to prove Proposition 4.1. Proof. There are several cases to be considered. The surface may be either separating or not, either orientable or not, either closed or not. The proofs of all those cases are similar, but some subtle differences may appear. So we write all the details for each case. Suppose the surface S has k boundary components, denoted by c1 , c2 , ..., ck . Different ci may be contained in the same torus component, but a torus component can contain at most 2 of such ci . In fact, suppose ci , cj ⊂ T , a torus component of ∂M . Now f |T is an orientation reversing involution on T . ci and cj are two parallel circles on T , pointwise fixed by f . It’s easy to see that f interchange the two connected components of T − ci ∪ cj , thus T cannot contain any other component cl other than ci and cj . Without loss of generality, we can assume that among the boundary components of S, c2i−1 and c2i are in the same torus boundary component, i = 1, 2, ..., m. and c2m+j , j = 1, 2, ..., l are contained in other l different boundary components. Case (1): S is an orientable surface of genus g. Note rk π1 (S) = 2g + k − 1 if k > 0 and
rk π1 (S) = 2g
if k = 0.
(4.1)
We will divide the discussion into two subcases according to whether S is separating or not. (i) S is separating. Then it is easy to see that k = 2m and l = 0. Cutting M along the surface S, we get two homeomorphic components M1 , M2 , and f interchanges them. Now each pair c2i−1 , c2i bounds an annulus in M1 connecting S, which increases the genus of S by 1. So we obtained a boundary component of M1 with genus (g + k2 ). If k > 0, then k > 2g + k − 1 = rk π1 (S) . 2 rk π1 (M ) ≥ 2 rk π1 (M1 ) ≥ 2 rk H1 (M1 , Q) ≥ 2 g + 2 The first and the third inequalities and the last equality are based on Lemma 4.2(2), (3) and (4.1) respectively. If k = 0, then M1 is a hyperbolic 3-manifold with a totally geodesic boundary component S, which is incompressible. Then 2 rk π1 (M ) ≥ 2 rk π1 (M1 ) > 2g = rk π1 (S) . Those two inequalities and one equality are based on Lemma 4.2(2), Lemma 4.3 (also Remark 4.4) and (4.1) respectively. (ii) S is non-separating. Cutting M along S, we get a new connected manifold M with two copies of S, denoted by S1 and S2 , contained in ∂M . Suppose S1 , S2 are contained in the same boundary component S of M . Then k > 0 and S consist of S1 , S2 and 2m + l annulus, which is clearly closed and orientable, and g ∂M ≥ g S = 2g + 2m + l − 1 = 2g + k − 1.
(4.2)
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As before, by Lemma 4.2(2), (3) and (4.2) we get rk(H1 (S , Q)) = 2g + k − 1. rk π1 M ≥ rk H1 M , Q ≥ 2
(4.3)
Now f = f |M −S is an involution on M , which keeps the boundary component S invariant and interchanges S1 and S1 . Now let’s take two copies of M , denote them by M1 and M2 , and glue them on S1 and S2 via Let r be the reflection on M ˜ about S1 ∪ S2 . Then we have a free the identity to get a new manifold M. ˜ defined as involution f˜ on M f˜|M1 = f ◦ r
and f˜|M2 = r ◦ f .
→M /f˜ = M is a two fold covering. It is easy to verify that π : M Applying Lemma 4.2(1) for n = 2, Lemma 4.2(2), (4.3) and (4.1) (recall that in this case k > 0), we have ) + 1 ≥ rk π1 M + 1 ≥ 2g + k > rk π1 (S) . 2 rk π1 (M ) ≥ rk π1 (M Suppose S1 and S2 belong to two different components of ∂M . Then l = 0 and each component consists of one Si and m annuli, hence g ∂M ≥ g(S1 ) + m + g(S2 ) + m = 2g + k.
(4.4)
˜ → M using the Doubling two copies of M along S1 and S2 and constructing a 2-fold covering M involution f as before, by (4.4) we can prove similarly that: 2 rk π1 (M ) ≥ 2g + k + 1 > rk π1 (S) . Case (2): S is non-orientable surface of genus g (connected sum of g real projective planes). Note rk π1 (S) = g + k − 1 if k > 0 and
rk π1 (S) = g
if k = 0.
(4.5)
In this case S is non-separating. As before, we cut M along S to get a new manifold M with one boundary component S consisting of the orientable double cover S of S and 2m + l annuli, and we have g ∂M ≥ g S = g − 1 + 2m + l = k + g − 1.
(4.6)
The involution f = f |M −S provides a covering transformation of S → S. Again, we glue two copies of and a double covering M → M. M along S to get the manifold M If k > 0, as before, applying Lemma 4.2(1) for n = 2, Lemma 4.2(2), (3), (4.6) and (4.5) in order, we have
) + 1 ≥ rk π1 M + 1 ≥ rk H1 M , Q + 1 ≥ k + g > rk π1 (S) . 2 rk π1 (M ) ≥ rk π1 (M If k = 0, applying Lemma 4.2(1) for n = 2, Lemma 4.2(2), Lemma 4.3 (also Remark 4.4), (4.6) and (4.5) in order, we have ) + 1 ≥ rk π1 M + 1 > g − 1 + 1 = rk π1 (S) . 2 rk π1 (M ) ≥ rk π1 (M We finished the proof of Proposition 4.1. 2
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Acknowledgements The first author was partially supported by Beijing International Center of Mathematical Research. The second author was partially supported by grant No. 11071006 of the National Natural Science Foundation of China. The authors thank Ian Agol, David Gabai, Boju Jiang, and Hao Zheng for valuable communications and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]
M. Bestvina, M. Handel, Train tracks and automorphisms of free groups, Ann. Math. 135 (1992). J. Hempel, 3-Manifolds as viewed from the curve complex, Topology 40 (3) (2001) 631–657. J. Hempel, 3-Manifolds, Ann. of Math. Studies, vol. 86, Princeton University Press, 1976. B.J. Jiang, S.D. Wang, Q. Zhang, Bounds for fixed points and fixed subgroups on surfaces and graphs, Algebr. Geom. Topol. 11 (4) (2011) 2297–2318. S. Kojima, Deformations of hyperbolic 3-cone-manifolds, J. Differ. Geom. 49 (3) (1998) 469–516. D.D. Long, G.A. Niblo, Subgroup separability and 3-manifold groups, Math. Z. 207 (2) (1991) 209–215. C. Maclachlan, A. Reid, The Arithmetic of Hyperbolic 3-Manifolds, Graduate Texts in Mathematics, vol. 219, SpringerVerlag, New York, 2003. E. Moise, Geometric Topology in Dimensions 2 and 3, Graduate Texts in Mathematics, vol. 47, Springer-Verlag, New York, Heidelberg, 1977. P. Scott, T. Wall, Topological methods in group trheory, in: Homological Group Theory, Proc. Sympos., Durham, 1977, Cambridge Univ. Press, 1979, pp. 137–203. W. Thurston, Geometry and Topology of 3-Manifolds, Princeton University Lecture Notes, 1979–1981. W. Thurston, Three dimensional manifolds, Kleinian groups and hyperbolic geometry, Bull. Am. Math. Soc. 316 (1982) 357–381. Q. Zhou, The moduli space of hyperbolic cone structures, J. Differ. Geom. 51 (3) (1999) 517–550.
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Topology and its Applications www.elsevier.com/locate/topol
Fixed subgroups of automorphisms of hyperbolic 3-manifold groups Jianfeng Lin, Shicheng Wang ∗ Department of Mathematics, Peking University, Beijing, China
a r t i c l e
i n f o
Article history: Received 18 February 2014 Accepted 6 May 2014 Available online xxxx
a b s t r a c t For fixed subgroups Fix(φ) of automorphisms φ of hyperbolic 3-manifold groups π1 (M ), we observe that rk(Fix(φ)) < 2 rk(π1 (M )) and the constant 2 in the inequality is sharp; we also classify all possible groups Fix(φ). © 2014 Elsevier B.V. All rights reserved.
Keywords: Fixed subgroups Automorphisms Hyperbolic 3-manifold
1. Introduction For a group G and an automorphism φ : G → G, we define Fix(φ) = {ω ∈ G | φ(ω) = ω}, which is a subgroup of G, and use rk(G) to denote the rank of G. The so called Scott conjecture proved 20 years ago in a celebrate work of M. Bestvina and M. Handel [1] states that: Theorem 1.1. For each automorphism φ of a free group G = Fn , rk Fix(φ) ≤ rk(G). In a recent paper by B.J. Jiang, S.D. Wang and Q. Zhang [4], it is proved that Theorem 1.2. For each automorphisms φ of a compact surface group G = π1 (S), rk Fix(φ) ≤ rk(G). It is obvious that the bounds given in Theorem 1.1 and Theorem 1.2 are sharp and can be achieved by the identity maps. * Corresponding author. E-mail addresses: [email protected] (J. Lin), [email protected] (S. Wang). http://dx.doi.org/10.1016/j.topol.2014.05.020 0166-8641/© 2014 Elsevier B.V. All rights reserved.
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In this note, we will address the similar problem for hyperbolic 3-manifold groups. We call a compact 3-manifold M hyperbolic, if M is orientable, and the interior of M admits a complete hyperbolic structure of finite volume (then M is either closed or ∂M is a union of tori). In this case G = π1 (M ) is isomorphic to a cofinite volume torsion free Kleinian group. A main observation in this paper is the following: Theorem 1.3. For each automorphism φ of a hyperbolic 3-manifold group, G = π1 (M ), rk Fix(φ) < 2 rk(G), and the upper bound is sharp when G runs over all hyperbolic 3-manifold groups. Theorem 1.3 is a conclusion of the following Theorem 1.4 and Theorem 1.5. Theorem 1.4. There exists a sequence of automorphisms φn : π1 (Mn ) → π1 (Mn ) of closed hyperbolic 3-manifolds Mn such that Fix(φn ) is the group of a closed surface, and rk(Fix(φn )) >2− rk(π1 (Mn ))
as n → ∞
for any > 0. Theorem 1.5. Suppose φ is an automorphism of G = π1 (M ), where M is a hyperbolic 3-manifold. Then rk(Fix(φ)) < 2 rk(G). The proof of Theorem 1.4 is self-contained up to some primary (and elegant) facts on hyperbolic geometry and on combinatoric topology and group theory. Roughly speaking each (Mi , φi ) in Theorem 1.4 is constructed as follows: We first construct the hyperbolic 3-manifold Pi with connected totally geodesic boundary. Then we double two copies of Pi along their boundaries to get the closed hyperbolic 3-manifold Mi . The reflection of Mi alone ∂Pi induces an automorphism φi : π1 (Mi ) → π1 (Mi ) with Fix(φ) = π1 (∂Pi ). In this process all involved ranks are carefully controlled, we get the inequality in Theorem 1.4. To prove Theorem 1.5, besides some combinatorial arguments on topology and on group theory, we need the following Theorem 1.6 which classifies all possible groups Fix(φ) for automorphisms φ of hyperbolic 3-manifold groups. Recall that each automorphism φ of π1 (M ) can be realized by an isometry f on M according to Mostow rigidity theorem. Theorem 1.6. Suppose G = π1 (M ), where M is a hyperbolic 3-manifold, and φ is an automorphism of G. Then Fix(φ) is one of the following types: the whole group G; the trivial group {e}; Z; Z ⊕ Z; a surfaces group π1 (S), where S can be orientable or not, and closed or not. More precisely (1) Suppose φ is induced by an orientation preserving isometry. (i) Fix(φ) is either Z, or Z ⊕ Z, or G, or {e}; moreover (ii) If M is closed, then Fix(φ) is either Z or G; (2) Suppose φ is induced by an orientation reversing isometry f . (i) If φ2 = id, then Fix(φ) is either Z or {e}; (ii) If φ2 = id, then Fix(φ) is either {e}, or the surface group π1 (S), where S is an embedded surface in M that is pointwise fixed by f . Theorem 1.6 is proved by using the algebraic version of Mostow Rigidity theorem, as well as some hyperbolic geometry and covering space argument.
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Fig. 1. Truncated hyperbolic tetrahedron.
The paper is organized as follows: In Section 2 we will prove Theorem 1.4 generalizing the examples from closed hyperbolic 3-manifolds to hyperbolic 3-manifolds with cusps. Theorem 1.6 and Theorem 1.5 will be proved in Section 3 and Section 4 respectively. Suppose a compact 3-manifold M is hyperbolic and S is a proper embedded surface in M . We say S is totally geodesic surface if S o , the interior of S, is totally geodesic. With an abuse of notation, we will use the same notation M (S) to present the interior of M (S). For terminologies not defined, see [3] and [10] for geometry and topology of 3-manifolds, and see [9] for group theory. 2. Proof of Theorem 1.4 In this section, we construct examples stated in Theorem 1.4. As sketched in Section 1, the first step is to construct hyperbolic manifolds with boundaries and control the ranks of the fundamental groups. There are different approaches to construct such manifolds. We will use the most original and the most direct one due to Thurston. (For another approach see Remark 2.6.) In Thurston’s Lecture Notes (Section 3.2 of [10]), there is a very concrete and beautiful construction of hyperbolic 3-manifolds with totally geodesic boundaries involving primary hyperbolic geometry only. In 3-dimensional hyperbolic space H 3 , there is a one-parameter family of truncated hyperbolic tetrahedron as in Fig. 1: Each of its 8 faces is totally geodesic; each of its 18 edges is geodesic line segment. There are 4 triangle faces and 4 hexagon faces. The 12 edges of the 4 triangle faces have the same length, and the remaining 6 edges, we call them “inner edge”, also have the same length. The triangle faces are perpendicular to the hexagon faces. The angles between hexagon faces are all equal and can be arbitrary angles between 0◦ and 60◦ . We will use those polyhedra to construct hyperbolic manifolds with totally geodesic boundaries. Suppose we have some copies of tetrahedron. We pair the faces of tetrahedron and glue them together (therefore some edges and vertexes are also glued together). After that, we remove a neighborhood of the vertex and get a topological manifold P . A tetrahedron with its vertex neighborhood removed is homeomorphic to the truncated simplex mentioned above. Now suppose for some fixed k > 6, all k edges of the tetrahedron are glued together in the resulting complex. We can set the face angle α of the truncated simplex to be 2π k such that the hyperbolic structure of the truncated simplex fix together to give the hyperbolic structure of P , and the triangle faces of the truncated simplex are matched together to form the totally geodesic ∂P . It is easy to see that the number of vertexes of tetrahedrons (after gluing) equals the number of the boundary components.
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Fig. 2. New generators.
Moreover, if we remove the regular neighborhood of the inner edges in P , we will get a handlebody H. To see this, we remove the neighborhood of the 6 edges of a tetrahedron. Topologically, it is homeomorphic to D3 and the 4 tetrahedron faces are 4 disjoint disks on ∂D3 . Then, we glue them together. If we glue some 3-balls along disks on their boundaries, we get a handlebody. Therefore, P can be obtained by attaching m 2-handles on a handlebody of genus n + 1. It is easy to see that m is the number of inner edges after gluing and n is the number of tetrahedrons. Now we double P along its boundary to get a closed hyperbolic manifold DP. We have to control the rank of π1 (DP). This is done in the following lemma. Lemma 2.1. Suppose P is obtained by attaching l 2-handles to a handlebody of genus k. Then rk(π1 (DP)) ≤ k + l, where DP is the double of P . Proof. Suppose P is obtained from a handlebody H of genus k by attaching l 2-handles h1 , h2 , ..., hl with attaching curves γ1 , γ2 , ..., γl , where γ1 , γ2 , ..., γl are disjoint simple closed curves on ∂H, and for each 2-handle D2 × I, ∂D × I is identified with the attaching region N (γi ), the regular neighborhood of γi , for some i. Then we have P =H
{hi }.
{N (γi )}
Note that in the doubling DP, the two copies of handlebody H and H are glued together along ∂H − D2 × I of the 2-handle hi are glued alone the D2 × ∂I to get a solid torus i N (γi ), and each two copies Si , which is attached to H ∂H− N (γi ) H along the torus boundary formed by two copies of N (γi ) (see i Fig. 2). We have
DP =
H
∂H− i N (γi )
H
Si .
i
Because attaching solid tori along the torus boundaries of H ∂H− N (γi ) H does not increase the rank of i the fundamental group, we just need to control rk(π1 (H ∂H− N (γi ) H )). We consider the two skeleton of i this space. The two skeleton of the handlebody H consists of a surface of genus k and k copies of compressing disks. Therefore, the two-skeletons of H ∂H− N (γi ) H can be obtained as follows: starting from a surface i Sk of genus k, we glue two copies of Sk along Sk − i N (γi ). This is equal to attaching l copies of annulus along ∂N (γi ), i = 1, 2, ..., l. Then we glue k compressing disks on both sides.
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Fig. 3. Gluing tetrahedrons.
Comparing to the two-skeleton of H, we see that only l new generators are involved by the attached annulus and the new attaching disks do not increase the rank of fundamental group. Thus we get rk(π1 (H ∂H− N (γi ) H )) ≤ k + l. 2 i
Now we can construct our examples. We start from n (n > 3, 3 n) copies of the tetrahedron indicated in Fig. 3, where the edges are marked. We represent the faces by the edges around it. Each tetrahedron Ti has 4 faces (1, 3, 2)i , (4, 5, 3)i , (2, 6, 4)i , (5, 1, 6)i . Then we group the 4n faces into 2n pairs:
(1, 3, 2)i , (4, 5, 3)i+1 ;
(2, 6, 4)i , (5, 1, 6)i+1 ,
i = 1, 2, ..., n, and n + 1 ≡ 1.
The two faces in each pair are glued together, and the orders of the edges are preserved. (It’s easy to see that the arrows on the edges are preserved too.) Then we get a simplicial complex X. We write ai ↔ bj to indicate that the edge a in Ti is glued together with b in Tj . With this notation, we have 1k ↔ 4k+1 ↔ 6k+2 ↔ 1k+3 ;
2k ↔ 5k+1 ↔ 3k ↔ 2k−1 .
We first count the number of the edges after the gluing: Since 3 n, we see that the 3n edges 1∗ , 4∗ , 6∗ are glued together, and the 3n edges 2∗ , 5∗ , 3∗ are glued together, so there are two edges in X. Then we count the number of the vertices after the gluing: If we denote the initial point and the terminal point of the directed edge ik by I(ik ) and E(ik ), then we have: E(1k+1 ) ↔ I(3k+1 ) ↔ I(2k ) ↔ I(4k ) ↔ I(1k−1 ) ↔ E(2k−1 )
(2.1)
The first, third and fifth identifications are shown in Fig. 3. The second and fourth identifications follow from the facts that 3k+1 and 2k , 4k and 1k−1 are glued together as direct edges respectively. Since the two edges [1∗ ], [2∗ ] after the gluing. Eq. (2.1) implies that all ends of [1∗ ], [2∗ ] are identified to a point. Hence there is only one vertex in the simplicial complex X. Finally we check the orientation: If we use the right hand coordinate system, the faces (1, 3, 2) and (5, 1, 6) correspond to outward normal vectors while (2, 6, 4) and (4, 5, 3) correspond to inward ones. Since each inward face is glued with an outward one, the orientations are matched after gluing. Now if we remove a regular neighborhood of the unique vertex, by the discussion at the begin of this section, we get an orientable hyperbolic three manifold P with connected, totally geodesic boundary. As we have discussed, P can be constructed by attaching two 2-handles on a handle body of genus n + 1. The genus of ∂P is n − 1. Now double P along its boundary to get a closed 3-manifold DP. If we
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choose a point p ∈ ∂P as the base point, the reflection f on DP along ∂P will induce an automorphism φ : π1 (DP) → π1 (DP). The following lemma is a general fact about automorphisms that switch the two factors of the amalgamation of a group with itself. Lemma 2.2. In the construction above, Fix(φ) = Im(i∗ (π1 (∂P ))). Proof. Let G and G be two identical copies of π1 (P ) and H be their common subgroup π1 (∂P ) (recall that P is boundary incompressible). Then we have: π1 (DP) = G ∗H ∗G . For each right coset gi H of H in G, fix its representative gi . We choose the unit 1 as the representative for the right coset H itself. Then {gi H} give the right coset decomposition of H in G . According to [9, Theorem 1.7], each element γ in G ∗H ∗G can be written uniquely in a form γ = a1 b1 a2 b2 ...an bn h, where h ∈ H, ai is some representative gj and bi is some representative gk ; moreover ai = 1 implies i = 1 and bi = 1 implies i = n. Then φ(γ) = a1 b1 a2 b2 ...an bn h. By uniqueness of the standard form, it is direct to see that if φ(γ) = γ, then γ = h. hence only elements in H can be fixed by φ. 2 By Lemma 2.1, rk(π1 DP) ≤ n + 3. Since ∂P has genus n − 1, by Lemma 2.2 Fix(φ) = Im(i∗ (π1 (∂P ))) ∼ = π1 (∂P ) has rank 2n − 2. For each n > 2, construct such pair (DP, φ), and denoted as (Mn , φn ). Then 2n − 2 rk(Fix(φn )) ≥ > 2 − , rk(π1 (Mn )) n+3 for any > 0. Hence we finished the proof of Theorem 1.4.
as n → ∞
2
The construction in Theorem 1.4 for closed hyperbolic 3-manifold can be modified to the case of hyperbolic 3-manifold with cusps. Precisely: Proposition 2.3. There exists a sequence of hyperbolic 3-manifolds Mn with cusps and automorphisms φn : π1 (Mn ) → π1 (Mn ), such that Fix(φn ) is a free group, and rk(Fix(φn )) >2− rk(π1 (Mn ))
as n → ∞.
for any > 0. We give some theorems to prove Proposition 2.3. Theorem 2.4. ([11]) Suppose M is a hyperbolic 3 manifold with finite volume and f is an involution of M . Than M admits a hyperbolic structure with finite volume such that f is an isometry with respect to this structure. Theorem 2.5. ([5,12]) Suppose M is a hyperbolic 3-manifold with finite volume and α is a simple closed geodesic in M . Then M − α admits a hyperbolic structure with finite volume.
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Proof. In the proof of Theorem 1.4, the hyperbolic 3-manifold P with connected totally geodesic boundary is obtained by attaching two 2-handles to a handlebody H along the attaching curve γ1 and γ2 . Now we choose a simple non-separating closed geodesic α in ∂P such that α ⊂ ∂H \ N (γ1 ) ∪ N (γ2 ). Then α remains a geodesic in DP. Remove α from the closed hyperbolic manifold DP, we get a new hyperbolic manifold with a cusp by Theorem 2.5, denoted by DP . The reflection f on DP along ∂P defines an involution f on DP . By Theorem 2.4, DP admits a hyperbolic structure so that f is an isometry under this hyperbolic structure. As the fixed point set of an isometry, the non-closed surface ∂P − α must be totally geodesic, and therefore incompressible. If we chose the base point to be on ∂P − α and consider the automorphism φ induced by f , the same combinational group theory argument as before shows that Fix(φ) = π1 (∂P − α). Because ∂P has genus n − 1, n is the same as in the proof of Theorem 1.4, ∂P − α is two punctured surface of genus n − 2, hence π1 (∂P − α) is a free group of rank 2(n − 2) + 1 = 2n − 3. Now we control rk(π1 (DP )) via the same technique in the proof of Lemma 2.1: DP consist of two parts: the first part consists of two copies of the handlebody H glued along ∂H \ N (γ1 ) ∪ N (γ2 ) ∪ N (α); the second part consists of two solid tori resulting from the doubling of the 2-handles. The same argument as in the proof of Lemma 2.1 shows that rk(π1 (DP )) < n + 4, and we have 2n − 3 rk(Fix(φ)) > 2 − , ≥ n+4 rk(π1 (DP ))
as n → ∞,
for any > 0. 2 Remark 2.6. There is another way to construct Mn which we sketch as follows. It is based on a result on Heegaard Splitting: Theorem 2.7. ([2]) Let M be a closed oriented 3-manifold which is Seifert fibered or which contains an essential torus. Then any splitting of M is a Heegaard distance ≤ 2 splitting. Note that Theorem 2.7 in [2] is stated for closed 3-manifolds, but the argument there can be used to prove the similar theorem for non-closed case. Now consider a manifold M obtained by attaching a two handle to a handlebody of genus n (n > 1). Suppose that the distances in the curve complex between the attaching curve and the boundaries of the compressing disks of the handlebody are larger than 3. By [2], M contains no essential surface of genus smaller than 2. By Haken hyperbolization theorem in [11], M is a hyperbolic manifold with totally geodesic boundary. Let Mn be the double of M . By Lemma 2.1, the inequality in Theorem 1.4 can be proved as before. 3. Proof of Theorem 1.6 In this section, we will classify all the possible fixed subgroups of automorphisms of torsion free cofinite volume Kleinian groups. We use Iso(H3 ) (resp. Iso+ (H3 )) to denote the group of (resp. orientation preserving) isometries of the 3-dimensional hyperbolic 3-space. The most important tool is the following algebraic version of Mostow rigidity theorem. Most topologists know the geometric version of Mostow rigidity: Any homotopy equivalence between finite volume hyperbolic 3 manifolds can be homotopied to an isometry. The following algebraic version appears in [7], which is equivalent to the geometric version. Theorem 3.1. Let Γ1 and Γ2 be two cofinite volume Kleinian groups, and φ : Γ1 → Γ2 be an isomorphism between them. Then there exists γ ∈ Iso(H3 ) (γ may be orientation reversing) such that φ(α) = γαγ −1 for any α ∈ Γ1 .
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Now let’s prove Theorem 1.6. Proof. Since G = π1 (M ) is a hyperbolic 3-manifold group, G can be considered as cofinite volume torsion free Kleinian group in Iso+ (H3 ). Now G acts on H3 as the deck transformation group for the covering π : H3 → H3 /G ∼ = M . By Theorem 3.1, there exists γ ∈ Iso(H3 ) such that for any α ∈ G, φ(α) = γαγ −1 . Then Fix(φ) = {α ∈ G | αγ = γα}.
(3.1)
Because γGγ −1 = G, γ induces an isometry f of M such that the following diagram commutes. H3
γ
π
M
H3 π
f
(3.2)
M
The verification of Theorem 1.6 will be based on (3.1) and the verification will be divided into two cases according to whether γ in (3.1) is orientation preserving or not. Case (1): γ is orientation preserving. (i) If γ = e then clearly Fix(φ) = G. Below we assume that γ is nontrivial. It is well-known that each element in G is either hyperbolic or parabolic; moreover two nontrivial elements α, β in Iso+ (H3 ) commute if and only if one of the following cases happens: (a) Both α and β are parabolic elements and they share the same fixed point in the infinite sphere S ∞ . (b) Both α and β are non-parabolic elements (elliptic or hyperbolic) and they share the same axis. (c) Both α and β are elliptic elements with rotation angle π and their axes are perpendicular to each other. Since elements in G (therefore in Fix(φ)) cannot be elliptic, we just need to consider case (a) and case (b). In these two cases, if α, β, γ are all nontrivial, α commutes with β, β commutes with γ, then α commute with γ. We see that Fix(φ) is a torsion free abelian group. As we know, the fundamental group of a hyperbolic 3-manifold can contain torsion free abelian subgroups of ranks at most 2. We have proved that: if φ is induced by an orientation preserving map, Fix(φ) can only be e, Z, Z ⊕ Z, or G. (ii) If we further assume that M is closed, then we have more restrictions. First π1 (M ) contains no subgroup Z ⊕ Z for a closed hyperbolic 3-manifold M . Also we claim that Fix(φ) = {e}. In fact, the self-isometry of a closed hyperbolic 3 manifold is always periodic. So there exists positive integer n such that f n = id. By commuting diagram (3.2), γ n induces the identity on M , therefore γ n ∈ G. Then clearly γ n ∈ Fix(φ). If γ n = e, then Fix(φ) is not trivial. If γ n = e, then γ is an elliptic element, so there is an axis l pointwise fixed by γ. By the commuting diagram (3.2), π(l) is pointwise fixed by the isometry f . Since M is closed, the fixed point set of an orientation preserving isometry can only be a closed geodesic. So π(l) is a closed geodesic. This means that there is a hyperbolic covering transformation α ∈ G sharing the axis l with γ, so αγ = γα, and therefore α ∈ Fix(φ) by (3.1). We have actually proved that if φ is induced by an orientation preserving map and M is closed, then Fix(φ) can be either Z or G. Case (2): γ is orientation reversing. Note Fix(φ) ⊆ Fix(φ2 ) and φ2 is induced by an orientation preserving map. There are two subcases now: (i) φ2 = id. Then Fix(φ2 ) can only be e, Z or Z ⊕ Z by Case (1)(i) and its proof, therefore Fix(φ) can only be e, Z or Z ⊕ Z.
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But in fact the situation Z ⊕ Z never happens. Because in this situation Fix(φ2 ) = Z ⊕ Z, which was generated by two parabolic elements β1 and β2 sharing the same fixed point p on the infinite sphere. Since we assume that φ2 = id, γ 2 is also a parabolic element with the fixed point p. Since Fix(γ) ⊂ Fix(γ 2 ), one can derive that γ has the unique fixed point p in H3 ∪ S ∞ , and in the upper-half model of H3 (we set p = ∞), γ 2 , β1 , β2 are translations along some directions v1 , v2 , v respectively. Consider their extended action on the plane z = 0. Then γ acts as a conformal (orientation reversing) map on this plane (in order to see this, we can just compose γ with an arbitrary reflection r which fixes p to get an orientation preserving isometry. By classical fact, both r ◦ γ and r act conformally on plane z = 0, then so does γ). Because γ 2 is a translation, γ must act as an orientation reversing isometry on the plane z = 0. So γ|z=0 = r ◦ h, where h is a translation along the direction v and r a reflection along an invariant line of h. Then it is a direct verification that γ commutes with βi if and only if the directions of v and vi are either the same or opposite. But the directions of v1 and v2 are neither the same nor opposite, so γ cannot commute with both two generators of Z ⊕ Z. We have proved that in this subcase Fix(φ) is either e or Z. (ii) φ2 = id. Then γ 2 commute with the whole group G. Thus γ 2 = e and γ has order 2. An order 2 orientation reversing isometry of H 3 can only be the reflection along a single point or reflection along a geodesic plane. If γ is the reflection along a point p ∈ H3 , then p is the only fixed point of γ. For any α ∈ Fix(φ), we have γα = αγ by (3.1). Hence γα(p) = αγ(p) = α(p), that is γ also has fixed point α(p), hence p = α(p). Because α is a covering transformation, we must have α = e. We have proved that if γ is a reflection along a single point, then Fix(φ) is trivial. If γ is the reflection along a totally geodesic plane P . Then P is pointwise fixed by γ. Because of the commuting diagram (3.2), π(P ) is pointwise fixed by f . We know that the fixed point set of an orientation reversing isometry of a hyperbolic 3-manifold must be totally geodesic surfaces if it is dimension 2. Thus π(P ) S where S is a totally geodesic surface in M . (S may be non-orientable although M is orientable. And if M has cusps, S may have cusps too.) For any α ∈ Fix(φ), γα = αγ. Then for each x ∈ P , γα(x) = αγ(x) = α(x), that is α(x) ∈ P . It follows that P is invariant under α. Conversely, suppose a covering transformation α ∈ G such that α(P ) = P . Then it is easy to see that γα = αγ and therefore α ∈ Fix(φ). So Fix(φ) is exactly the covering transformations of the universal π|P covering map P −−−→ S. We have proved that Fix(φ) ∼ = π1 (S). 2 4. Proof of Theorem 1.5 By Theorem 1.6, the situation rk(Fix(φ)) > rk(π1 (M )) can appear only if Case (2)(ii) in Theorem 1.6 happens, and if Case (2)(ii) in Theorem 1.6 happens, then Fix(φ) = π1 (S) for some surface S which is pointwise fixed by an orientation reversing isometry f of order 2 on M . The proof of Theorem 1.5 will be completed by the following: Proposition 4.1. Suppose M is a hyperbolic 3-manifold and S is a proper embedded surface in M . If there is an orientation reversing isometry f of order 2 on M fixing S pointwisely. Then rk π1 (S) < 2 rk π1 (M ) . To prove Proposition 4.1, we need the following lemma which contains several elementary facts: Lemma 4.2. (1) Suppose G is a group with subgroup H of index n. Then rk(G) ≥
rk(H) + n − 1 . n
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(2) Suppose S is a boundary component of the compact 3-manifold M , and DS (M ) is the doubling of two copies of M along S. Then rk π1 DS (M ) ≥ rk π1 (M ) . (3) (Half lives, half dies Lemma) Suppose M is a compact orientable 3-manifold. Then
dim H 1 (∂M, Q) dim image i∗ : H 1 (∂M, Q) → H 1 (M, Q) = 2 where i∗ is induced by the inclusion i : ∂M → M . (4) Suppose M is a compact orientable 3-manifold and S is an incompressible boundary component of M . If the homomorphism π1 (S) → π1 (M ) induced by the inclusion is not a surjection, then there is a finite ˜ → M such that p−1 (S) contains more then one component. covering p : M Proof. (1) If rk(G) = k, we can find a 2-dimensional CW complex X with fundamental group G so that X ˜ be the n-sheet covering space corresponding to the subgroup H. has only one vertex and k edges. Let X ˜ and this lifted CW complex of X ˜ provides a presentation of Then there are n vertexes and nk edges in X, H with n(k − 1) + 1 generators. So rk(H) ≤ n(k − 1) + 1, that is rk(G) = k ≥
rk(H) + n − 1 . n
(2) It is clear that there is a reflection f about S on DS (M ), which provides a folding map DS (M ) → DS (M )/f ∼ = M , and which obviously induce an epimorphism between fundamental groups. Hence rk π1 (M ) ≥ rk π1 (M1 ) . (3) See [8, Section 23]. (4) By the assumptions, π1 (S) is a proper subgroup of π1 (M ) (we pick a base point x on S for both π1 (S) and π1 (M )). Pick a non-zero element α of π1 (M ) but not in π1 (S). By [6, Theorem 1] (peripheral subgroups are separable), there is a finite index subgroup H of π1 (M ) which contains π1 (S) but does not ˜ → M corresponding to H, then there is a component S˜ of contain α. Consider the finite covering p : M −1 p (S) homeomorphic to S (since π1 (S) ⊂ H), and p−1 (S) has more than one component (since α does not in H, the lift α ˜ of α with one end in S˜ must have another end in another component of p−1 (S)). 2 Lemma 4.3. Suppose M is a compact orientable 3-manifold and S is an incompressible boundary component of M with genus g. Then rk(π1 (M )) > g. Proof. Since S is incompressible, the homomorphism i∗ : π1 (S) → π1 (M ) induced by the inclusion is injective. Suppose i∗ is also surjective. Then rk(π1 (M )) = 2g and the inequality is trivial. Now assume i∗ is not surjective. Let p : M → M be the n sheet covering provided by the proof of Lemma 4.2(4). Then the preimage of S has m > 1 component. Now we can compute the sum of the genus of these m boundary components. Because S has Euler number 2 − 2g, the sum of the Euler number of the preimage of S is n(2 − 2g). Therefore, the sum of their genus is n(g − 1) + m. Since H 1 (∂M , Q) is a direct sum of the homology of the boundary components, it contains at least 2n(g − 1) + 2m copies of Q. By Lemma 4.2(3), we have rk(H 1 (M , Q)) ≥ n(g −1) +m. Since H 1 (M , Q) is a quotient group of π1 (M ), we have rk(π1 (M )) ≥ n(g − 1) + m. Since π1 (M ) is subgroup of π1 (M ) of index n, by Lemma 4.1(1) we have n(g − 1) + m + n − 1 m−1 =g+ . rk π1 (M ) ≥ n n Since m > 1. Then Lemma 4.3 is proved. 2
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Remark 4.4. Suppose M is a closed hyperbolic 3-manifold and S is a closed surface embedded in M which is pointwise fixed by an orientation reversing involution on M . Cutting M along S, we get a compact 3-manifold M (may be not connected) with a boundary component S. In the proof of Theorem 4.1 we will apply Lemma 4.3 to S ⊂ M directly by the following reason: Using the fact that S is a proper embedded surface in M which is pointwise fixed by an orientation reversing involution f on M and M contains no essential spheres and essential tori, it follows that S must be boundary incompressible (otherwise M would contain essential spheres). Now we start to prove Proposition 4.1. Proof. There are several cases to be considered. The surface may be either separating or not, either orientable or not, either closed or not. The proofs of all those cases are similar, but some subtle differences may appear. So we write all the details for each case. Suppose the surface S has k boundary components, denoted by c1 , c2 , ..., ck . Different ci may be contained in the same torus component, but a torus component can contain at most 2 of such ci . In fact, suppose ci , cj ⊂ T , a torus component of ∂M . Now f |T is an orientation reversing involution on T . ci and cj are two parallel circles on T , pointwise fixed by f . It’s easy to see that f interchange the two connected components of T − ci ∪ cj , thus T cannot contain any other component cl other than ci and cj . Without loss of generality, we can assume that among the boundary components of S, c2i−1 and c2i are in the same torus boundary component, i = 1, 2, ..., m. and c2m+j , j = 1, 2, ..., l are contained in other l different boundary components. Case (1): S is an orientable surface of genus g. Note rk π1 (S) = 2g + k − 1 if k > 0 and
rk π1 (S) = 2g
if k = 0.
(4.1)
We will divide the discussion into two subcases according to whether S is separating or not. (i) S is separating. Then it is easy to see that k = 2m and l = 0. Cutting M along the surface S, we get two homeomorphic components M1 , M2 , and f interchanges them. Now each pair c2i−1 , c2i bounds an annulus in M1 connecting S, which increases the genus of S by 1. So we obtained a boundary component of M1 with genus (g + k2 ). If k > 0, then k > 2g + k − 1 = rk π1 (S) . 2 rk π1 (M ) ≥ 2 rk π1 (M1 ) ≥ 2 rk H1 (M1 , Q) ≥ 2 g + 2 The first and the third inequalities and the last equality are based on Lemma 4.2(2), (3) and (4.1) respectively. If k = 0, then M1 is a hyperbolic 3-manifold with a totally geodesic boundary component S, which is incompressible. Then 2 rk π1 (M ) ≥ 2 rk π1 (M1 ) > 2g = rk π1 (S) . Those two inequalities and one equality are based on Lemma 4.2(2), Lemma 4.3 (also Remark 4.4) and (4.1) respectively. (ii) S is non-separating. Cutting M along S, we get a new connected manifold M with two copies of S, denoted by S1 and S2 , contained in ∂M . Suppose S1 , S2 are contained in the same boundary component S of M . Then k > 0 and S consist of S1 , S2 and 2m + l annulus, which is clearly closed and orientable, and g ∂M ≥ g S = 2g + 2m + l − 1 = 2g + k − 1.
(4.2)
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As before, by Lemma 4.2(2), (3) and (4.2) we get rk(H1 (S , Q)) = 2g + k − 1. rk π1 M ≥ rk H1 M , Q ≥ 2
(4.3)
Now f = f |M −S is an involution on M , which keeps the boundary component S invariant and interchanges S1 and S1 . Now let’s take two copies of M , denote them by M1 and M2 , and glue them on S1 and S2 via Let r be the reflection on M ˜ about S1 ∪ S2 . Then we have a free the identity to get a new manifold M. ˜ defined as involution f˜ on M f˜|M1 = f ◦ r
and f˜|M2 = r ◦ f .
→M /f˜ = M is a two fold covering. It is easy to verify that π : M Applying Lemma 4.2(1) for n = 2, Lemma 4.2(2), (4.3) and (4.1) (recall that in this case k > 0), we have ) + 1 ≥ rk π1 M + 1 ≥ 2g + k > rk π1 (S) . 2 rk π1 (M ) ≥ rk π1 (M Suppose S1 and S2 belong to two different components of ∂M . Then l = 0 and each component consists of one Si and m annuli, hence g ∂M ≥ g(S1 ) + m + g(S2 ) + m = 2g + k.
(4.4)
˜ → M using the Doubling two copies of M along S1 and S2 and constructing a 2-fold covering M involution f as before, by (4.4) we can prove similarly that: 2 rk π1 (M ) ≥ 2g + k + 1 > rk π1 (S) . Case (2): S is non-orientable surface of genus g (connected sum of g real projective planes). Note rk π1 (S) = g + k − 1 if k > 0 and
rk π1 (S) = g
if k = 0.
(4.5)
In this case S is non-separating. As before, we cut M along S to get a new manifold M with one boundary component S consisting of the orientable double cover S of S and 2m + l annuli, and we have g ∂M ≥ g S = g − 1 + 2m + l = k + g − 1.
(4.6)
The involution f = f |M −S provides a covering transformation of S → S. Again, we glue two copies of and a double covering M → M. M along S to get the manifold M If k > 0, as before, applying Lemma 4.2(1) for n = 2, Lemma 4.2(2), (3), (4.6) and (4.5) in order, we have
) + 1 ≥ rk π1 M + 1 ≥ rk H1 M , Q + 1 ≥ k + g > rk π1 (S) . 2 rk π1 (M ) ≥ rk π1 (M If k = 0, applying Lemma 4.2(1) for n = 2, Lemma 4.2(2), Lemma 4.3 (also Remark 4.4), (4.6) and (4.5) in order, we have ) + 1 ≥ rk π1 M + 1 > g − 1 + 1 = rk π1 (S) . 2 rk π1 (M ) ≥ rk π1 (M We finished the proof of Proposition 4.1. 2
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Acknowledgements The first author was partially supported by Beijing International Center of Mathematical Research. The second author was partially supported by grant No. 11071006 of the National Natural Science Foundation of China. The authors thank Ian Agol, David Gabai, Boju Jiang, and Hao Zheng for valuable communications and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]
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